# 能量均分定理

## 基本概念及简易例子

### 平移能量与理想气体

${\displaystyle H^{\mathrm {kin} }={\tfrac {1}{2}}m|\mathbf {v} |^{2}={\tfrac {1}{2}}m\left(v_{x}^{2}+v_{y}^{2}+v_{z}^{2}\right),}$

${\displaystyle v_{\mathrm {rms} }={\sqrt {\langle v^{2}\rangle }}={\sqrt {\frac {3k_{B}T}{m}}}={\sqrt {\frac {3RT}{M}}},}$

### 旋转能量与溶液中的分子滚翻

${\displaystyle H^{\mathrm {rot} }={\tfrac {1}{2}}(I_{1}\omega _{1}^{2}+I_{2}\omega _{2}^{2}+I_{3}\omega _{3}^{2}),}$

### 势能与谐波振荡器

${\displaystyle H^{\mathrm {pot} }={\tfrac {1}{2}}aq^{2},\,}$

${\displaystyle H=H^{\mathrm {kin} }+H^{\mathrm {pot} }={\frac {p^{2}}{2m}}+{\frac {1}{2}}aq^{2}.}$

${\displaystyle \langle H\rangle =\langle H^{\mathrm {kin} }\rangle +\langle H^{\mathrm {pot} }\rangle ={\tfrac {1}{2}}k_{B}T+{\tfrac {1}{2}}k_{B}T=k_{B}T,}$

### 粒子的淀积

${\displaystyle H^{\mathrm {grav} }=m_{b}gz\,,}$

## 能量均分定理的通用公式化

${\displaystyle \!{\Bigl \langle }x_{m}{\frac {\partial H}{\partial x_{n}}}{\Bigr \rangle }=\delta _{mn}k_{B}T.}$

1. 对所有n${\displaystyle {\Bigl \langle }x_{n}{\frac {\partial H}{\partial x_{n}}}{\Bigr \rangle }=k_{B}T}$
2. 对所有mn${\displaystyle {\Bigl \langle }x_{m}{\frac {\partial H}{\partial x_{n}}}{\Bigr \rangle }=0}$

${\displaystyle k_{B}T={\Bigl \langle }x_{n}{\frac {\partial H}{\partial x_{n}}}{\Bigr \rangle }=2\langle a_{n}x_{n}^{2}\rangle ,}$

${\displaystyle {\Bigl \langle }p_{k}{\frac {\partial H}{\partial p_{k}}}{\Bigr \rangle }={\Bigl \langle }q_{k}{\frac {\partial H}{\partial q_{k}}}{\Bigr \rangle }=k_{B}T.}$

${\displaystyle {\Bigl \langle }p_{k}{\frac {dq_{k}}{dt}}{\Bigr \rangle }=-{\Bigl \langle }q_{k}{\frac {dp_{k}}{dt}}{\Bigr \rangle }=k_{B}T.}$

${\displaystyle {\Bigl \langle }q_{j}{\frac {\partial H}{\partial q_{k}}}{\Bigr \rangle },\quad {\Bigl \langle }q_{j}{\frac {\partial H}{\partial p_{k}}}{\Bigr \rangle },\quad {\Bigl \langle }p_{j}{\frac {\partial H}{\partial p_{k}}}{\Bigr \rangle },\quad {\Bigl \langle }p_{j}{\frac {\partial H}{\partial q_{k}}}{\Bigr \rangle },\quad {\Bigl \langle }q_{k}{\frac {\partial H}{\partial p_{k}}}{\Bigr \rangle },}$   及   ${\displaystyle {\Bigl \langle }p_{k}{\frac {\partial H}{\partial q_{k}}}{\Bigr \rangle }}$

j≠k则皆为零。

### 与均功定理的关系

${\displaystyle {\Bigl \langle }\sum _{k}q_{k}{\frac {\partial H}{\partial q_{k}}}{\Bigr \rangle }={\Bigl \langle }\sum _{k}p_{k}{\frac {\partial H}{\partial p_{k}}}{\Bigr \rangle }={\Bigl \langle }\sum _{k}p_{k}{\frac {dq_{k}}{dt}}{\Bigr \rangle }=-{\Bigl \langle }\sum _{k}q_{k}{\frac {dp_{k}}{dt}}{\Bigr \rangle },}$

## 应用

### 理想气体定律

{\displaystyle {\begin{aligned}\langle H^{\mathrm {kin} }\rangle &={\frac {1}{2m}}\langle p_{x}^{2}+p_{y}^{2}+p_{z}^{2}\rangle \\&={\frac {1}{2}}{\biggl (}{\Bigl \langle }p_{x}{\frac {\partial H^{\mathrm {kin} }}{\partial p_{x}}}{\Bigr \rangle }+{\Bigl \langle }p_{y}{\frac {\partial H^{\mathrm {kin} }}{\partial p_{y}}}{\Bigr \rangle }+{\Bigl \langle }p_{z}{\frac {\partial H^{\mathrm {kin} }}{\partial p_{z}}}{\Bigr \rangle }{\biggr )}={\frac {3}{2}}k_{B}T\end{aligned}}}

{\displaystyle {\begin{aligned}\langle \mathbf {q} \cdot \mathbf {F} \rangle &={\Bigl \langle }q_{x}{\frac {dp_{x}}{dt}}{\Bigr \rangle }+{\Bigl \langle }q_{y}{\frac {dp_{y}}{dt}}{\Bigr \rangle }+{\Bigl \langle }q_{z}{\frac {dp_{z}}{dt}}{\Bigr \rangle }\\&=-{\Bigl \langle }q_{x}{\frac {\partial H}{\partial q_{x}}}{\Bigr \rangle }-{\Bigl \langle }q_{y}{\frac {\partial H}{\partial q_{y}}}{\Bigr \rangle }-{\Bigl \langle }q_{z}{\frac {\partial H}{\partial q_{z}}}{\Bigr \rangle }=-3k_{B}T,\end{aligned}}}

${\displaystyle 3Nk_{B}T=-{\biggl \langle }\sum _{k=1}^{N}\mathbf {q} _{k}\cdot \mathbf {F} _{k}{\biggr \rangle }}$

${\displaystyle -{\biggl \langle }\sum _{k=1}^{N}\mathbf {q} _{k}\cdot \mathbf {F} _{k}{\biggr \rangle }=P\oint _{\mathrm {surface} }\mathbf {q} \cdot \mathbf {dS} ,}$

${\displaystyle {\boldsymbol {\nabla }}\cdot \mathbf {q} ={\frac {\partial q_{x}}{\partial q_{x}}}+{\frac {\partial q_{y}}{\partial q_{y}}}+{\frac {\partial q_{z}}{\partial q_{z}}}=3,}$

${\displaystyle P\oint _{\mathrm {surface} }\mathbf {q} \cdot \mathbf {dS} =P\int _{\mathrm {volume} }\left({\boldsymbol {\nabla }}\cdot \mathbf {q} \right)dV=3PV,}$

${\displaystyle 3Nk_{B}T=-{\biggl \langle }\sum _{k=1}^{N}\mathbf {q} _{k}\cdot \mathbf {F} _{k}{\biggr \rangle }=3PV,}$

${\displaystyle PV=Nk_{B}T=nRT,\,}$

### 双原子气体

${\displaystyle H={\frac {\left|\mathbf {p} _{1}\right|^{2}}{2m_{1}}}+{\frac {\left|\mathbf {p} _{2}\right|^{2}}{2m_{2}}}+{\frac {1}{2}}aq^{2},}$

### 极度相对性理想气体

${\displaystyle H^{\mathrm {kin} }\approx cp=c{\sqrt {p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}}.}$

${\displaystyle p_{x}{\frac {\partial H^{\mathrm {kin} }}{\partial p_{x}}}=c{\frac {p_{x}^{2}}{\sqrt {p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}}}}$

{\displaystyle {\begin{aligned}\langle H^{\mathrm {kin} }\rangle &={\biggl \langle }c{\frac {p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}{\sqrt {p_{x}^{2}+p_{y}^{2}+p_{z}^{2}}}}{\biggr \rangle }\\&={\Bigl \langle }p_{x}{\frac {\partial H^{\mathrm {kin} }}{\partial p_{x}}}{\Bigr \rangle }+{\Bigl \langle }p_{y}{\frac {\partial H^{\mathrm {kin} }}{\partial p_{y}}}{\Bigr \rangle }+{\Bigl \langle }p_{z}{\frac {\partial H^{\mathrm {kin} }}{\partial p_{z}}}{\Bigr \rangle }\\&=3k_{B}T\end{aligned}}}

### 非理想气体

${\displaystyle \langle h^{\mathrm {pot} }\rangle =\int _{0}^{\infty }4\pi r^{2}\rho U(r)g(r)\,dr}$

${\displaystyle H=\langle H^{\mathrm {kin} }\rangle +\langle H^{\mathrm {pot} }\rangle ={\frac {3}{2}}Nk_{B}T+2\pi N\rho \int _{0}^{\infty }r^{2}U(r)g(r)\,dr}$

${\displaystyle 3Nk_{B}T=3PV+2\pi N\rho \int _{0}^{\infty }r^{3}U'(r)g(r)\,dr}$

### 非谐振器

${\displaystyle H^{\mathrm {pot} }=Cq^{s}}$

${\displaystyle k_{B}T={\Bigl \langle }q{\frac {\partial H^{\mathrm {pot} }}{\partial q}}{\Bigr \rangle }=\langle q\cdot sCq^{s-1}\rangle =\langle sCq^{s}\rangle =s\langle H^{\mathrm {pot} }\rangle }$

${\displaystyle H^{\mathrm {pot} }=\sum _{n=2}^{\infty }C_{n}q^{n}}$

${\displaystyle k_{B}T={\Bigl \langle }q{\frac {\partial H^{\mathrm {pot} }}{\partial q}}{\Bigr \rangle }=\sum _{n=2}^{\infty }\langle q\cdot nC_{n}q^{n-1}\rangle =\sum _{n=2}^{\infty }nC_{n}\langle q^{n}\rangle }$

${\displaystyle \langle H^{\mathrm {pot} }\rangle ={\frac {1}{2}}k_{B}T-\sum _{n=3}^{\infty }\left({\frac {n-2}{2}}\right)C_{n}\langle q^{n}\rangle }$

### 布朗运动

${\displaystyle {\frac {d\mathbf {v} }{dt}}={\frac {1}{m}}\mathbf {F} =-{\frac {\mathbf {v} }{\tau }}+{\frac {1}{m}}\mathbf {F} ^{\mathrm {rnd} },}$

${\displaystyle {\Bigl \langle }\mathbf {r} \cdot {\frac {d\mathbf {v} }{dt}}{\Bigr \rangle }+{\frac {1}{\tau }}\langle \mathbf {r} \cdot \mathbf {v} \rangle =0}$

（由于Frnd跟位置向量r不相关）。使用数学恒等式

${\displaystyle {\frac {d}{dt}}\left(\mathbf {r} \cdot \mathbf {r} \right)={\frac {d}{dt}}\left(r^{2}\right)=2\left(\mathbf {r} \cdot \mathbf {v} \right)}$

${\displaystyle {\frac {d}{dt}}\left(\mathbf {r} \cdot \mathbf {v} \right)=v^{2}+\mathbf {r} \cdot {\frac {d\mathbf {v} }{dt}},}$

${\displaystyle {\frac {d^{2}}{dt^{2}}}\langle r^{2}\rangle +{\frac {1}{\tau }}{\frac {d}{dt}}\langle r^{2}\rangle =2\langle v^{2}\rangle ={\frac {6}{m}}k_{B}T,}$

${\displaystyle \langle H^{\mathrm {kin} }\rangle ={\Bigl \langle }{\frac {p^{2}}{2m}}{\Bigr \rangle }=\langle {\tfrac {1}{2}}mv^{2}\rangle ={\tfrac {3}{2}}k_{B}T}$

${\displaystyle \langle r^{2}\rangle ={\frac {6k_{B}T\tau ^{2}}{m}}\left(e^{-t/\tau }-1+{\frac {t}{\tau }}\right)}$

${\displaystyle \langle r^{2}\rangle \approx {\frac {3k_{B}T}{m}}t^{2}=\langle v^{2}\rangle t^{2}}$

${\displaystyle \langle r^{2}\rangle \approx {\frac {6k_{B}T\tau }{m}}t=6\gamma k_{B}Tt}$

### 恒星物理学

${\displaystyle H_{\mathrm {tot} }^{\mathrm {grav} }=-\int _{0}^{R}{\frac {4\pi r^{2}G}{r}}M(r)\,\rho (r)\,dr,}$

${\displaystyle H_{\mathrm {tot} }^{\mathrm {grav} }=-{\frac {3GM^{2}}{5R}},}$

${\displaystyle H_{\mathrm {tot} }^{\mathrm {grav} }=-{\frac {3GM^{2}}{5R}},}$

${\displaystyle {\Bigl \langle }r{\frac {\partial H^{\mathrm {grav} }}{\partial r}}{\Bigr \rangle }=\langle -H^{\mathrm {grav} }\rangle =k_{B}T={\frac {3GM^{2}}{5RN}}.}$

### 恒星的形成

${\displaystyle {\frac {3GM^{2}}{5R}}>3Nk_{B}T}$

${\displaystyle M={\frac {4}{3}}\pi R^{3}\rho }$

${\displaystyle M_{J}^{2}=\left({\frac {5k_{B}T}{Gm_{p}}}\right)^{3}\left({\frac {3}{4\pi \rho }}\right)}$

## 推导

### 动能与麦克斯韦-波兹曼分布

${\displaystyle f(v)=4\pi \left({\frac {m}{2\pi k_{B}T}}\right)^{3/2}\!\!v^{2}\exp {\Bigl (}{\frac {-mv^{2}}{2k_{B}T}}{\Bigr )}}$

${\displaystyle \langle H^{\mathrm {kin} }\rangle =\langle {\tfrac {1}{2}}mv^{2}\rangle =\int _{0}^{\infty }{\tfrac {1}{2}}mv^{2}\ f(v)\ dv={\tfrac {3}{2}}k_{B}T}$

### 二次能量与配分函数

${\displaystyle Z_{x}=\int _{-\infty }^{\infty }dx\ e^{-\beta Ax^{2}}={\sqrt {\frac {\pi }{\beta A}}}}$

${\displaystyle \langle H_{x}\rangle =-{\frac {\partial \log Z_{x}}{\partial \beta }}={\frac {1}{2\beta }}={\frac {1}{2}}k_{B}T}$

### 一般证明

${\displaystyle d\Gamma =\prod _{i}dq_{i}dp_{i}}$

${\displaystyle \Gamma (E,\Delta E)=\int _{H\in \left[E,E+\Delta E\right]}d\Gamma .}$

${\displaystyle \Sigma (E)=\int _{H

${\displaystyle \int _{H\in \left[E,E+\Delta E\right]}\ldots d\Gamma =\Delta E{\frac {\partial }{\partial E}}\int _{H

${\displaystyle \Gamma =\Delta E\ {\frac {\partial \Sigma }{\partial E}}=\Delta E\ \rho (E),}$

${\displaystyle {\frac {1}{T}}={\frac {\partial S}{\partial E}}=k_{b}{\frac {\partial \log \Sigma }{\partial E}}=k_{b}{\frac {1}{\Sigma }}\,{\frac {\partial \Sigma }{\partial E}}}$

#### 正则系综

${\displaystyle {\mathcal {N}}\int e^{-\beta H(p,q)}d\Gamma =1,}$

${\displaystyle {\mathcal {N}}\int \left[e^{-\beta H(p,q)}x_{k}\right]_{x_{k}=a}^{x_{k}=b}d\Gamma _{k}+{\mathcal {N}}\int e^{-\beta H(p,q)}x_{k}\beta {\frac {\partial H}{\partial x_{k}}}d\Gamma =1,}$

${\displaystyle {\mathcal {N}}\int e^{-\beta H(p,q)}x_{k}{\frac {\partial H}{\partial x_{k}}}\,d\Gamma ={\Bigl \langle }x_{k}{\frac {\partial H}{\partial x_{k}}}{\Bigr \rangle }={\frac {1}{\beta }}=k_{B}T.}$

#### 微正则系综

{\displaystyle {\begin{aligned}{\Bigl \langle }x_{m}{\frac {\partial H}{\partial x_{n}}}{\Bigr \rangle }&={\frac {1}{\Gamma }}\,\int _{H\in \left[E,E+\Delta E\right]}x_{m}{\frac {\partial H}{\partial x_{n}}}\,d\Gamma \\&={\frac {\Delta E}{\Gamma }}\,{\frac {\partial }{\partial E}}\int _{H

{\displaystyle {\begin{aligned}\int _{H

${\displaystyle {\Bigl \langle }x_{m}{\frac {\partial H}{\partial x_{n}}}{\Bigr \rangle }=\delta _{mn}{\frac {1}{\rho }}\,{\frac {\partial }{\partial E}}\int _{H

${\displaystyle {\Bigl \langle }x_{m}{\frac {\partial H}{\partial x_{n}}}{\Bigr \rangle }=\delta _{mn}{\Bigl (}{\frac {1}{\Sigma }}{\frac {\partial \Sigma }{\partial E}}{\Bigr )}^{-1}=\delta _{mn}{\Bigl (}{\frac {\partial \log \Sigma }{\partial E}}{\Bigr )}^{-1}=\delta _{mn}k_{B}T}$

${\displaystyle \!{\Bigl \langle }x_{m}{\frac {\partial H}{\partial x_{n}}}{\Bigr \rangle }=\delta _{mn}k_{B}T,}$

## 限制

### 量子效应引起的失败

${\displaystyle P(E_{n})={\frac {e^{-n\beta h\nu }}{Z}}}$

${\displaystyle Z=\sum _{n=0}^{\infty }e^{-n\beta h\nu }={\frac {1}{1-e^{-\beta h\nu }}}}$

${\displaystyle \langle H\rangle =\sum _{n=0}^{\infty }E_{n}P(E_{n})={\frac {1}{Z}}\sum _{n=0}^{\infty }nh\nu \ e^{-n\beta h\nu }=-{\frac {1}{Z}}{\frac {\partial Z}{\partial \beta }}=-{\frac {\partial \log Z}{\partial \beta }}}$

Z的式子代入得最后结果[7]

${\displaystyle \langle H\rangle =h\nu {\frac {e^{-\beta h\nu }}{1-e^{-\beta h\nu }}}}$

## 注释及参考资料

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## 延伸阅读

• Khinchin, AI. Mathematical Foundations of Statistical Mechanics (G. Gamow, translator). New York: Dover Publications. 1949: pp. 93–98. ISBN 978-0-486-63896-6.
• Mohling, F. Statistical Mechanics: Methods and Applications. John Wiley and Sons. 1982: pp. 137–139, 270–273, 280, 285–292. ISBN 978-0-470-27340-1.
• Tolman, RC. Statistical Mechanics, with Applications to Physics and Chemistry. Chemical Catalog Company. 1927: pp. 72–81. ASIN B00085D6OO
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